A particle undergoes three successive displac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the unit vector towards East be $\hat{i}$ and towards North be $\hat{j}$.$s_1 = \sqrt{2} \cos 45^{\circ} \hat{i} + \sqrt{2} \sin 45^{\circ} \h

Vedclass · Accepted Answer

$\sqrt{14 - 4\sqrt{3}} \ m$

Vedclass · Answer

(B) Let the unit vector towards East be $\hat{i}$ and towards North be $\hat{j}$.$s_1 = \sqrt{2} \cos 45^{\circ} \hat{i} + \sqrt{2} \sin 45^{\circ} \hat{j} = \hat{i} + \hat{j}$$s_2 = -2 \hat{j}$ (since it is due south)$s_3 = 4 \cos 150^{\circ} \hat{i} + 4 \sin 150^{\circ} \hat{j} = 4(-\frac{\sqrt{3}}{2}) \hat{i} + 4(\frac{1}{2}) \hat{j} = -2\sqrt{3} \hat{i} + 2 \hat{j}$Net displacement $\vec{s} = s_1 + s_2 + s_3 = (1 - 2\sqrt{3}) \hat{i} + (1 - 2 + 2) \hat{j} = (1 - 2\sqrt{3}) \hat{i} + \hat{j}$Magnitude $|\vec{s}| = \sqrt{(1 - 2\sqrt{3})^2 + (1)^2} = \sqrt{1 + 12 - 4\sqrt{3} + 1} = \sqrt{14 - 4\sqrt{3}} \ m$.

$A$ particle undergoes three successive displacements given by $s_1 = \sqrt{2} \ m$ north-east,$s_2 = 2 \ m$ due south,and $s_3 = 4 \ m$ at $30^{\circ}$ north of west. The magnitude of the net displacement is:

Similar Questions

Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $(\vec A + \vec B)$ is $n$ times the magnitude of $(\vec A - \vec B)$. The angle between $\vec A$ and $\vec B$ is

$A$ vector $\vec Q$ which has a magnitude of $8$ is added to the vector $\vec P$ which lies along the $x$-axis. The resultant of the two vectors lies along the $y$-axis and has a magnitude twice that of $\vec P$. The magnitude of $\vec P$ is:

$A$ body moves due East with a velocity of $20 \, km/h$ and then due North with a velocity of $15 \, km/h$. The resultant velocity is .......... $km/h$.

Which of the following relations is true for two unit vectors $\hat{A}$ and $\hat{B}$ making an angle $\theta$ to each other?

The angle $\beta$ between vector $\overrightarrow{A}$ and the resultant vector $(\overrightarrow{A}-\overrightarrow{B})$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module